Proposition 2. Let w,u be paths in Q. If w is compatible (http://planetmath.org/PathAlgebraOfAQuiver) with u then F⁢(w) is compatible (http://planetmath.org/PathAlgebraOfAQuiver) with F⁢(u). The inverseimplication holds if and only if F0 is an injective function.

Now assume that w, u are paths, which are not compatible (http://planetmath.org/PathAlgebraOfAQuiver). If F0 is injective, then by proposition 2 F⁢(w) and F⁢(u) are also not compatible (http://planetmath.org/PathAlgebraOfAQuiver) and thus

F¯⁢(w⋅u)=F¯⁢(0)=0=F¯⁢(w)⋅F¯⁢(u).

On the other hand, if F0 is not injective, then there are paths w, u which are not compatible (http://planetmath.org/PathAlgebraOfAQuiver), but F⁢(w), F⁢(u) are. Assume, that F¯ is a homomorphism of algebras. Then

0=F¯⁢(0)=F¯⁢(w⋅u)=F¯⁢(w)⋅F¯⁢(u)≠0

because of the compatibility (http://planetmath.org/PathAlgebraOfAQuiver). The contradiction shows that F¯ is not a homomorphism of algebras. This completes the proof. □